# Solution

Show that the product of any three consecutive integers is divisible by 6.

Consider three consecutive integers: $n$, $n+1$, and $n+2$. We wish to show that their product, $n(n+1)(n+2)$ is divisible by 6. Suppose we break this task into two pieces. It would be sufficient to show that their product is both even and divisible by 3.

To show that $n(n+1)(n+2)$ is even, note that $n$ itself must be even or odd.

• If $n$ is even, then $n(n+1)(n+2)$ is the product of two even numbers and an odd number, which must then be even.
• If $n$ is odd, then $n(n+1)(n+2)$ is the product of two odd numbers and an even number, which again, must be even.

Thus, regardless of the integer $n$ chosen, $n(n+1)(n+2)$ must be even.

It remains to show that $n(n+1)(n+2)$ is divisible by 3.

Consider the possible remainders that can be left when $n$ is divided by 3:

• If $n$ divided by 3 leaves a remainder of 0, then $3 \mid n$.
• If $n$ divided by 3 leaves a remainder of 1, then $3 \mid n+2$.
• If $n$ divided by 3 leaves a remainder of 2, then $3 \mid n+1$.

In each case, some factor of $n(n+1)(n+2)$ is divisible by 3. As such, regardless of the integer $n$ chosen, $n(n+1)(n+2)$ is divisible by 3.

We have shown that $n(n+1)(n+2)$ is both even and divisible by 3. Thus, the product of any three consecutive integers is divisible by 6.